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For a derived category of a Noetherian ring (or perhaps more generally), can we talk about a Periodicity Theorem? We have Thick Subcategory Theorems and Nilpotence Theorems (HPS 91) for D(R), and in some more general cases, but I haven't seen, in my limited review of the literature, a theorem describing how such data might "stratify" the endomorphisms of the unit object (in this case I guess R). Or, in a similar vein, finding specific maps whose cofiber lies in perhaps the "next higher" level. Perhaps this is connected to the fact that for a ring, unlike in the case of finite p-local spectra, there isn't really a linear ordering of primes, necessarily. Anyone know of any work on this idea? Or perhaps know a straightforward answer?

Thanks!

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    $\begingroup$ "isn't really a linear ordering of the primes" Exactly. The periodicity theorem comes from the stratification of the moduli stack of formal groups, in the case of spectra. For D(R) there isn't really an analog in general. It's not clear to me what such a "periodicity theorem" would even say, in this case... But I could just be silly! $\endgroup$
    – Dylan Wilson
    Commented Nov 1, 2012 at 23:14
  • $\begingroup$ What I mean is, there are analogues of the Morava K-theories, which are basically the residue fields $k(\mathfrak{p})$. These things homology acyclics (use Tor), in a sensible way, and seem to give a sensible thick subcategory theorem. The question for me is, can I find a self-map $f$ of a perfect complex $X$ of type $\mathfrak{p}$ whose cofiber is type (and this isn't really defined) $\mathfrak{p}+1$, but I'd like it to be some kind of minimal deformation of $\mathfrak{p}$ or something. Whatever that means. And I would like for this to induce periodic families in $[R,R]$. $\endgroup$
    – Jonathan Beardsley
    Commented Nov 1, 2012 at 23:18
  • $\begingroup$ And I'm not sure I agree that the periodicity theorem "comes from" that stratification. But if you can ascertain that, then you're right, we shouldn't have something analogous in rings necessarily, unless we can find a similar structure. $\endgroup$
    – Jonathan Beardsley
    Commented Nov 1, 2012 at 23:19

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